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(C) Let the foot of the perpendicular from the origin $O(0,0)$ to the line $PQ$ be $R(h, k)$.Since $OR \perp PQ$,the slope of $OR$ is $m_1 = \frac{k}{

Vedclass · Accepted Answer

$(x^2 + y^2)^2 \left( \frac{1}{x^2} + \frac{1}{y^2} \right) = 4a^2$

Vedclass · Answer

(C) Let the foot of the perpendicular from the origin $O(0,0)$ to the line $PQ$ be $R(h, k)$.Since $OR \perp PQ$,the slope of $OR$ is $m_1 = \frac{k}{h}$.The slope of line $PQ$ is $m_2 = -\frac{h}{k}$.The equation of line $PQ$ passing through $(h, k)$ is $y - k = -\frac{h}{k}(x - h)$,which simplifies to $hx + ky = h^2 + k^2$.The intercepts of this line on the axes are $Q\left( \frac{h^2 + k^2}{h}, 0 ight)$ and $P\left( 0, \frac{h^2 + k^2}{k} ight)$.Since $O, P, Q$ lie on a circle of radius $a$,the diameter of the circle is the hypotenuse $PQ$ of the right-angled triangle $	riangle OPQ$.The length $PQ = \sqrt{\left( \frac{h^2 + k^2}{h} ight)^2 + \left( \frac{h^2 + k^2}{k} ight)^2} = 2a$.Squaring both sides: $(h^2 + k^2)^2 \left( \frac{1}{h^2} + \frac{1}{k^2} ight) = 4a^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 \left( \frac{1}{x^2} + \frac{1}{y^2} ight) = 4a^2$.

$A$ circle of constant radius $a$ passes through the origin $O$ and cuts the coordinate axes at points $P$ and $Q$. The equation of the locus of the foot of the perpendicular from $O$ to $PQ$ is:

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